operation on single elements in MATLAB

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Noriham B
Noriham B 2022 年 7 月 29 日
コメント済み: Noriham B 2022 年 7 月 29 日
greeting all the experts,
x=-2:1:2; %coordinates x
n=length(x);
y=x.^2; %coordinates y
z=[x;y];
from the above example, I will have matrix z (2x5)=[-2 -1 0 1 2: 4 1 0 1 4]. This will give 5 point on the graph which are (-2,4), (-1,1), (0,0),(1,1) (2,4). then, my next step is, i want to find the distance for each points.
distance 1 from (-2,4) to (-1,1)
distance 2 from (-1,1) to (0,0)
distance 3 from (0,0) to(1,1)
distance 4 from (1,1) to (2,4)
My problem/question, how to type the distance formula generally to conduct operation on each elements?
distance formula = sqrt((x2-x1)^2+(y2-y1)^2)

採用された回答

Matt J
Matt J 2022 年 7 月 29 日
編集済み: Matt J 2022 年 7 月 29 日
x=-2:1:2; %coordinates x
n=length(x);
y=x.^2; %coordinates y
z=[x;y]
z = 2×5
-2 -1 0 1 2 4 1 0 1 4
interDistances=vecnorm(diff(z,1,2),2,1) %the result
interDistances = 1×4
3.1623 1.4142 1.4142 3.1623
  4 件のコメント
Walter Roberson
Walter Roberson 2022 年 7 月 29 日
vecnorm needs r2017b .
We assume you have a new enough version of MATLAB as you did not enter a release when you created your question.
Noriham B
Noriham B 2022 年 7 月 29 日
noted Prof/Dr./Sir Walter. I will try to get the newer version after this. Thanks :)

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その他の回答 (1 件)

Chunru
Chunru 2022 年 7 月 29 日
x=-2:1:2; %coordinates x
y=x.^2; %coordinates y
z=[x;y]
z = 2×5
-2 -1 0 1 2 4 1 0 1 4
d = diff(z, 1, 2) % diff along 2nd dim
d = 2×4
1 1 1 1 -3 -1 1 3
d = vecnorm(d) % distance
d = 1×4
3.1623 1.4142 1.4142 3.1623
  1 件のコメント
Noriham B
Noriham B 2022 年 7 月 29 日
Dear Prof/Dr./Sir Chunru. Tqvm for the help. U also use vecnorm like Matt J. I will study more about vecnorm for future need. Tq also to MATLAB for giving us opportunity to ask for help in this platform.

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